Proof step 1 from proof countevn-bounded-ind (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  n ≤ # a, m ≤ n, ¬ countevn(a, m, n, p) ≤ (even(m) ⊃ (n + 1 - m) / 2;(n - m) / 2) ⊦

Rule argument

Additional information

Proof step 2 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 3:

  m ≤ n0 +1, n0 +1 ≤ # a, ¬ countevn(a, m, n0 +1, p) ≤ (even(m) ⊃ (n0 +1 + 1 - m) / 2;(n0 +1 - m) / 2), ∀ a, m, p. ¬(n0 ≤ # a ∧ m ≤ n0 ∧ ¬ countevn(a, m, n0, p) ≤ (even(m) ⊃ (n0 + 1 - m) / 2;(n0 - m) / 2)) ⊦

Rule argument

Additional information

Used simplifier rules: countevn-base: m ≤ # a ⊦ countevn(a, m, m, p) = 0; nat freely generated by 0, +1;; ⊦ m ≤ 0 ↔ m = 0; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ 0 ≤ m; (s)

Proof step 3 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  a ≠ [], (even(m) ⊃ (2 + n0 - m) / 2;(n0 + 1 - m) / 2) < countevn(a, m, n0 + 1, p), n0 < # a, m ≤ # a, m ≤ n0 + 1, ∀ a0, m0, p0. ¬ n0 ≤ # a0 ∨ ¬ m0 ≤ n0 ∨ countevn(a0, m0, n0, p0) ≤ (even(m0) ⊃ ((n0 - m0) + 1) / 2;(n0 - m0) / 2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s) ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic le-one: ⊦ m ≤ n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m +1 = m + 1; (s) ⊦ m + 1 ≤ n ↔ m < n; (s)

Proof step 4 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 5:

  a ≠ [], n0 < # a, (even(m) ⊃ (2 + n0 - m) / 2;(n0 + 1 - m) / 2) < countevn(a, m, n0 + 1, p), m ≤ n0 + 1, m ≤ # a, ¬ n0 ≤ # a ∨ ¬ m ≤ n0 ∨ countevn(a, m, n0, p) ≤ (even(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2) ⊦

Rule argument

Additional information

Used terms: a, m, p

Proof step 5 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  a ≠ [], (even(m) ⊃ (2 + n0 - m) / 2;(n0 + 1 - m) / 2) < countevn(a, m, n0 + 1, p), n0 < # a, m ≤ # a, m ≤ n0 + 1, ¬ m ≤ n0 ∨ countevn(a, m, n0, p) ≤ (even(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)

Proof step 6 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  m ≤ n, n ≤ # a ⊦ countevn(a, m, n + 1, p) = countevn(a, m, n, p) + (even(n) ∧ p(a[n]) ⊃ 1;0)

• Proof step 8:

  a ≠ [], n0 < # a, (even(m) ⊃ (2 + n0 - m) / 2;(n0 + 1 - m) / 2) < countevn(a, m, n0, p) + (even(n0) ∧ p(a[n0]) ⊃ 1;0), m ≤ n0 + 1, m ≤ # a, m ≤ n0, n0 ≤ # a, ¬ m ≤ n0 ∨ countevn(a, m, n0, p) ≤ (even(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2) ⊦

Rule argument

Additional information

Used simplifier rules: countevn-base: m ≤ # a ⊦ countevn(a, m, m, p) = 0; ⊦ m < n → m ≤ n; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s) eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m + 1 ≠ 0; (s) ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) ⊦ even(n + 1) ↔ odd(n); (s) ⊦ m + n - n + n1 = m - n1; (s) ⊦ m - m = 0; (s) ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: a, m, n0, p

Proof step 7 from proof countevn-bounded-ind (previous step) (back to tree)

Lemma countevn-rec

Proof step 8 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  a ≠ [], m < # a, (even(m) ⊃ (n0 - m) / 2 + 1;((n0 - m) + 1) / 2) < countevn(a, m, n0, p) + (even(n0) ∧ p(a[n0]) ⊃ 1;0), n0 < # a, countevn(a, m, n0, p) ≤ (even(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2), m ≤ n0 + 1, m ≤ n0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s) ⊦ (m + 2) / 2 = m / 2 + 1; (s)

Proof step 9 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 10:

  a ≠ [], n0 < # a, m < # a, (n0 - m) / 2 + 1 < countevn(a, m, n0, p) + (even(n0) ∧ p(a[n0]) ⊃ 1;0), even(m), m ≤ n0, m ≤ n0 + 1, countevn(a, m, n0, p) ≤ (even(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2) ⊦

• Proof step 18:

  a ≠ [], n0 < # a, m < # a, ((n0 - m) + 1) / 2 < countevn(a, m, n0, p) + (even(n0) ∧ p(a[n0]) ⊃ 1;0), m ≤ n0, m ≤ n0 + 1, countevn(a, m, n0, p) ≤ (even(m) ⊃ ((n0 - m) + 1) / 2;(n0 - m) / 2), ¬ even(m) ⊦

Rule argument

Additional information

Proof step 10 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  a ≠ [], (n0 - m) / 2 + 1 < countevn(a, m, n0, p) + (even(n0) ∧ p(a[n0]) ⊃ 1;0), m < # a, n0 < # a, even(m), m ≤ n0 + 1, m ≤ n0, countevn(a, m, n0, p) * 2 ≤ (n0 - m) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; (s)

Proof step 11 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 12:

  n0 = n1 + m, a ≠ [], n0 < # a, m < # a, n1 / 2 + 1 < countevn(a, m, n0, p) + (even(n0) ∧ p(a[n0]) ⊃ 1;0), even(m), countevn(a, m, n0, p) * 2 ≤ n1 + 1, m ≤ n0, m ≤ n0 + 1 ⊦

Rule argument

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, m, n1

Proof step 12 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  a ≠ [], m < # a, n1 + m < # a, n1 / 2 + 1 < countevn(a, m, n1 + m, p) + (even(n1) ∧ p(a[n1 + m]) ⊃ 1;0), even(m), countevn(a, m, n1 + m, p) * 2 ≤ n1 + 1 ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ even(m) → (even(m + n) ↔ even(n)); (s) ⊦ m ≤ m + n; (s)

Proof step 13 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 14:

  n1 = m0 * 2 + n0, 2 ≠ 0, a ≠ [], m0 + 1 < countevn(a, m, n1 + m, p) + (even(n1) ∧ p(a[n1 + m]) ⊃ 1;0), n1 + m < # a, m < # a, n0 < 2, even(m), countevn(a, m, n1 + m, p) * 2 ≤ n1 + 1 ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n1, 2, m0, n0

Proof step 14 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  a ≠ [], n0 < 2, m < # a, (m0 * 2 + n0) + m < # a, m0 + 1 < countevn(a, m, (m0 * 2 + n0) + m, p) + (even(m0 * 2 + n0) ∧ p(a[(m0 * 2 + n0) + m]) ⊃ 1;0), even(m), countevn(a, m, (m0 * 2 + n0) + m, p) * 2 ≤ (m0 * 2 + n0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 15 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 16:

  n0 = 0, a ≠ [], m0 + 1 < countevn(a, m, (m0 * 2 + n0) + m, p) + (even(m0 * 2 + n0) ∧ p(a[(m0 * 2 + n0) + m]) ⊃ 1;0), (m0 * 2 + n0) + m < # a, m < # a, n0 < 2, even(m), countevn(a, m, (m0 * 2 + n0) + m, p) * 2 ≤ (m0 * 2 + n0) + 1 ⊦

Rule argument

Additional information

Used simplifier rules: countevn-rec: m ≤ n, n ≤ # a ⊦ countevn(a, m, n + 1, p) = countevn(a, m, n, p) + (even(n) ∧ p(a[n]) ⊃ 1;0); nat freely generated by 0, +1;; ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; (s) ⊦ m < 1 ↔ m = 0; (s) ⊦ m ≤ m + n; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ¬ m + n < m; (s) ⊦ even(m) → (even(m + n) ↔ even(n)); (s) ⊦ even(n + 1) ↔ odd(n); (s) ⊦ m +1 = m + 1; (s) ⊦ ¬ odd(2 * m); (s) ⊦ even(2 * m); (s) ⊦ n + 0 = n; (s) ⊦ m * n ≤ m1 * n + n ↔ m ≤ m1 + 1 ∨ n = 0; (s) ⊦ m < n → (n ≤ m ↔ false); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult

Proof step 16 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  a ≠ [], m0 + 1 < countevn(a, m, m0 * 2 + m, p) + (p(a[m0 * 2 + m]) ⊃ 1;0), m0 * 2 + m < # a, m < # a, even(m), countevn(a, m, m0 * 2 + m, p) ≤ m0 ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + 1 ≠ 0; (s) ⊦ ¬ m + n < m; (s) ⊦ m < n → n ≠ 0; (s) fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ n + 0 = n; (s) mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult ⊦ even(2 * m); (s) ⊦ 2 * n ≤ 2 * m + 1 ↔ n ≤ m; (s)

Proof step 17 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule case distinction

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m < n → n ≠ 0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ m ≤ m0 ∧ n ≤ n0 → ¬ m0 + n0 < m + n; (s)

Proof step 18 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  a ≠ [], ((n0 - m) + 1) / 2 < countevn(a, m, n0, p) + (even(n0) ∧ p(a[n0]) ⊃ 1;0), m < # a, n0 < # a, countevn(a, m, n0, p) * 2 + m < # a, odd(m), m ≤ n0 + 1, m ≤ n0, countevn(a, m, n0, p) * 2 + m ≤ n0 ⊦

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ even(n) ↔ ¬ odd(n); (ss) n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; (s) ⊦ m ≤ n → (n < m ↔ false); (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) m0 ≠ 0 ⊦ n ≤ m / m0 ↔ n * m0 ≤ m; (s)

Proof step 19 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 20:

  n0 = n1 + m, a ≠ [], countevn(a, m, n0, p) * 2 + m < # a, n0 < # a, m < # a, (n1 + 1) / 2 < countevn(a, m, n0, p) + (even(n0) ∧ p(a[n0]) ⊃ 1;0), odd(m), countevn(a, m, n0, p) * 2 + m ≤ n0, m ≤ n0, m ≤ n0 + 1 ⊦

Rule argument

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n0, m, n1

Proof step 20 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  a ≠ [], (n1 + 1) / 2 < countevn(a, m, n1 + m, p) + (even(n1 + m) ∧ p(a[n1 + m]) ⊃ 1;0), m < # a, n1 + m < # a, countevn(a, m, n1 + m, p) * 2 + m < # a, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1 ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s) ⊦ m ≤ m + n; (s)

Proof step 21 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

• Proof step 22:

  n1 + 1 = m0 * 2 + n0, 2 ≠ 0, a ≠ [], countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, m0 < countevn(a, m, n1 + m, p) + (even(n1 + m) ∧ p(a[n1 + m]) ⊃ 1;0), n0 < 2, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1 ⊦

Rule argument

Additional information

Used simplifier rules: n ≠ 0 ⊦ m0 = m / n ∧ n0 = m % n ↔ m = m0 * n + n0 ∧ n0 < n; (elim) Used terms: n1 + 1, 2, m0, n0

Proof step 22 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 23:

  n1 + 1 = m0 * 2 + n0, a ≠ [], n0 < 2, m0 < countevn(a, m, n1 + m, p) + (even(n1 + m) ∧ p(a[n1 + m]) ⊃ 1;0), m < # a, n1 + m < # a, countevn(a, m, n1 + m, p) * 2 + m < # a, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1 ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 23 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 24:

  n1 + 1 = m0 * 2 + n0, n0 = 0, a ≠ [], countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, m0 < countevn(a, m, n1 + m, p) + (even(n1 + m) ∧ p(a[n1 + m]) ⊃ 1;0), n0 < 2, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1 ⊦

• Proof step 44:

  n1 + 1 = m0 * 2 + n0, n0 = n +1, a ≠ [], countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, m0 < countevn(a, m, n1 + m, p) + (even(n1 + m) ∧ p(a[n1 + m]) ⊃ 1;0), n0 < 2, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1 ⊦

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;;

Proof step 24 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  n1 + 1 = m0 * 2, a ≠ [], m0 < countevn(a, m, n1 + m, p) + (even(n1 + m) ∧ p(a[n1 + m]) ⊃ 1;0), m < # a, n1 + m < # a, countevn(a, m, n1 + m, p) * 2 + m < # a, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1 ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n + 0 = n; (s)

Proof step 25 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 26:

  n1 + 1 = m0 * 2, a ≠ [], countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, m0 < countevn(a, m, n1 + m, p) + 1, even(n1 + m), odd(m), p(a[n1 + m]), countevn(a, m, n1 + m, p) * 2 ≤ n1 ⊦

• Proof step 32:

  n1 + 1 = m0 * 2, a ≠ [], countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, m0 < countevn(a, m, n1 + m, p) + 0, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1, ¬ even(n1 + m) ∨ ¬ p(a[n1 + m]) ⊦

Rule argument

Additional information

Proof step 26 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  n1 + 1 = m0 * 2, a ≠ [], m < # a, n1 + m < # a, countevn(a, m, n1 + m, p) * 2 + m < # a, even(n1 + m), odd(m), p(a[n1 + m]), countevn(a, m, n1 + m, p) * 2 ≤ n1, m0 ≤ countevn(a, m, n1 + m, p) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s)

Proof step 27 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 28:

  ⊦ even(m + n) ↔ even(m) ∧ even(n) ∨ odd(m) ∧ odd(n)

• Proof step 29:

  n1 + 1 = m0 * 2, a ≠ [], countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, odd(m), p(a[n1 + m]), m0 ≤ countevn(a, m, n1 + m, p), countevn(a, m, n1 + m, p) * 2 ≤ n1, even(n1) ∧ even(m) ∨ odd(n1) ∧ odd(m) ⊦

Rule argument

Additional information

Used terms: n1, m

Proof step 28 from proof countevn-bounded-ind (previous step) (back to tree)

A spec-lemma

Proof step 29 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  n1 + 1 = m0 * 2, a ≠ [], m < # a, n1 + m < # a, countevn(a, m, n1 + m, p) * 2 + m < # a, odd(m), odd(n1), p(a[n1 + m]), countevn(a, m, n1 + m, p) * 2 ≤ n1, m0 ≤ countevn(a, m, n1 + m, p) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ odd(n) → ¬ even(n); (s)

Proof step 30 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 31:

  ⊦ odd(n) ↔ (∃ m. n = m + m + 1)

Rule argument

Additional information

Used simplifier rules: countevn-rec: m ≤ n, n ≤ # a ⊦ countevn(a, m, n + 1, p) = countevn(a, m, n, p) + (even(n) ∧ p(a[n]) ⊃ 1;0); ⊦ m * n + n = m1 * n ↔ m + 1 = m1 ∨ n = 0; (s) ⊦ 2 * n ≤ 2 * m + 1 ↔ n ≤ m; (s) ⊦ n0 < m0 → (m0 * m + n0 < m0 * n ↔ m < n); (s) ⊦ 2 * m < 2 * n ↔ m < n; (s) mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult ⊦ m + m = 2 * m; (s) ⊦ m ≤ m + n; (s) ⊦ m ≤ n ↔ ¬ n < m; (ss) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic ⊦ m < n → n ≠ 0; (s) ⊦ m +1 = m + 1; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ ¬ m + n < m; (s) ⊦ m + 1 ≤ n ↔ m < n; (s) ⊦ m ≤ n → (n < m ↔ false); (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1

Proof step 31 from proof countevn-bounded-ind (previous step) (back to tree)

A spec-lemma

Proof step 32 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 33:

  n1 + 1 = m0 * 2, a ≠ [], m0 < countevn(a, m, n1 + m, p), countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1, ¬ even(n1 + m) ∨ ¬ p(a[n1 + m]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n + 0 = n; (s)

Proof step 33 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 34:

  ⊦ even(m + n) ↔ even(m) ∧ even(n) ∨ odd(m) ∧ odd(n)

• Proof step 35:

  n1 + 1 = m0 * 2, a ≠ [], m < # a, n1 + m < # a, countevn(a, m, n1 + m, p) * 2 + m < # a, m0 < countevn(a, m, n1 + m, p), odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1, ¬(even(n1) ∧ even(m) ∨ odd(n1) ∧ odd(m)) ∨ ¬ p(a[n1 + m]) ⊦

Rule argument

Additional information

Used terms: n1, m

Proof step 34 from proof countevn-bounded-ind (previous step) (back to tree)

A spec-lemma

Proof step 35 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 36:

  n1 + 1 = m0 * 2, a ≠ [], m0 < countevn(a, m, n1 + m, p), countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1, ¬ odd(n1) ∨ ¬ p(a[n1 + m]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ odd(n) → ¬ even(n); (s)

Proof step 36 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 37:

  n1 + 1 = m0 * 2, a ≠ [], m < # a, n1 + m < # a, countevn(a, m, n1 + m, p) * 2 + m < # a, m0 < countevn(a, m, n1 + m, p), odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1, ¬ odd(n1) ⊦

• Proof step 40:

  n1 + 1 = m0 * 2, a ≠ [], m < # a, n1 + m < # a, countevn(a, m, n1 + m, p) * 2 + m < # a, m0 < countevn(a, m, n1 + m, p), odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1, ¬ p(a[n1 + m]) ⊦

Rule argument

Additional information

Proof step 37 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  n1 + 1 = m0 * 2, a ≠ [], m0 < countevn(a, m, n1 + m, p), countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, even(n1), odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ odd(n) ↔ ¬ even(n); (ss)

Proof step 38 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 39:

  ⊦ even(n) ↔ (∃ m. m + m = n)

Rule argument

Additional information

Used simplifier rules: ⊦ m + m = 2 * m; (s) ⊦ n0 < n → (m * n + n0 = m0 * n ↔ m = m0 ∧ n0 = 0); (ws) mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: n1

Proof step 39 from proof countevn-bounded-ind (previous step) (back to tree)

A spec-lemma

Proof step 40 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 41:

  n1 + 1 = m0 * 2, m0 = m1 +1, a ≠ [], m0 < countevn(a, m, n1 + m, p), countevn(a, m, n1 + m, p) * 2 + m < # a, n1 + m < # a, m < # a, odd(m), countevn(a, m, n1 + m, p) * 2 ≤ n1, ¬ p(a[n1 + m]) ⊦

Rule argument

Additional information

Used simplifier rules: nat freely generated by 0, +1;; mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic

Proof step 41 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  a ≠ [], m1 + 1 < countevn(a, m, 2 * m1 + m, p) + (even(2 * m1 + m) ∧ p(a[2 * m1 + m]) ⊃ 1;0), (countevn(a, m, 2 * m1 + m, p) + (even(2 * m1 + m) ∧ p(a[2 * m1 + m]) ⊃ 1;0)) * 2 + m < # a, (2 * m1 + 1) + m < # a, m < # a, odd(m), countevn(a, m, 2 * m1 + m, p) + (even(2 * m1 + m) ∧ p(a[2 * m1 + m]) ⊃ 1;0) ≤ m1, ¬ p(a[(2 * m1 + 1) + m]) ⊦

Rule argument

Additional information

Used simplifier rules: countevn-rec: m ≤ n, n ≤ # a ⊦ countevn(a, m, n + 1, p) = countevn(a, m, n, p) + (even(n) ∧ p(a[n]) ⊃ 1;0); mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ 2 * n ≤ 2 * m + 1 ↔ n ≤ m; (s) ⊦ ¬ m + n < m; (s) ⊦ m + n < m + n0 ↔ n < n0; (s) ⊦ m +1 = m + 1; (s) ⊦ m < n → n ≠ 0; (s) trans-nat: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m ≤ n ↔ ¬ n < m; (ss) ⊦ m ≤ m + n; (s) ⊦ m ≠ 0 → (m * n = 0 ↔ n = 0); (s) ⊦ m < m + n ↔ n ≠ 0; (s) ⊦ 1 < 2 * n ↔ n ≠ 0; (s) n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s) ⊦ m * (n + 1) = m * n + m; (s)

Proof step 42 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 43:

  ⊦ even(m + n) ↔ even(m) ∧ even(n) ∨ odd(m) ∧ odd(n)

Rule argument

Additional information

Used simplifier rules: ⊦ even(2 * m); (s) ⊦ odd(n) → ¬ even(n); (s) ⊦ ¬ odd(2 * m); (s) mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult ⊦ n + 0 = n; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat ⊦ m < n → n ≠ 0; (s) ⊦ ¬ m + n < m; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: 2 * m1, m

Proof step 43 from proof countevn-bounded-ind (previous step) (back to tree)

A spec-lemma

Proof step 44 from proof countevn-bounded-ind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 45:

  countevn(a, m, m0 * 2 + m, p) = m0, a ≠ [], m0 * 2 + m < # a, m < # a, even(m0 * 2 + m), odd(m), p(a[m0 * 2 + m]) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m; (s) ⊦ m ≤ n → (n ≤ m ↔ m = n); (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ m < m + n ↔ n ≠ 0; (s) fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat ⊦ n + 0 = n; (s) mult-c: ⊦ m * n = n * m; from specification nat-mult mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult 0 ≠ m ⊦ (* m) * n ≤ m * n0 ↔ n ≤ n0; (s) ⊦ m < 1 ↔ m = 0; (s) n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n; (s) ⊦ m +1 = m + 1; (s) ⊦ m + n = m + n0 ↔ n = n0; (s)

Proof step 45 from proof countevn-bounded-ind (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 46:

  ⊦ even(m) → (even(m + n) ↔ even(n))

Rule argument

Additional information

Used simplifier rules: ⊦ even(n) → ¬ odd(n); (s) ⊦ even(2 * m); (s) mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult mult-c: ⊦ m * n = n * m; from specification nat-mult Used terms: m0 * 2, m

Proof step 46 from proof countevn-bounded-ind (previous step) (back to tree)

A spec-lemma